Which number(s) is(are) equal to the quarter of its (their) square?
SolutionIf x is the number to find, then the quarter of its square is (1/4) x2.
x is equal to the quarter of its square hence: x = (1/4) x2First write the above equation with its right side equal to zero.
x - (1/4) x2 = 0
x(1 - (1/4)x) = 0 , factor x out
solutions: x = 0 and x = 4
Check answers.
1) x = 0 , quarter of its square is (1/4) 02 = 0 . Hence x and the quarter of its square are equal.
2) x = 4 , the quarter of its square is (1/4) 42 = 4 . Hence x and the quarter of its square are equal.

A car travels from A to B at a speed of 40 mph then returns, using the same road, from B to A at a speed of 60 mph. What is the average speed for the round trip?
SolutionThe average speed is given by.
total distance / total time
If x is the distance from A to B, then the total distance is equal to 2x (away and return). The total T time is equal to time t1 = x / 40 from A to B (away) plus t2 = x / 60 from B to A (return). Hence
T = x / 40 + x / 60 = 100 x / 2400 = x / 24
The average speed is given by
2x / (x/24) = 48 mph

Tom travels 60 miles per hour going to a neighboring city and 50 miles per hour coming back using the same road. He drove a total of 5 hours away and back. What is the distance from Tom's house to the city he visited?(round your answer to the nearest mile).
SolutionLet x be the distance traveled to the city, then the time taken is given by
distance / speed = x / 60
and the time to return is given by
distance / speed = x / 50
The total time to travel and return is 5 hours. Hence
x / 60 + x / 50 = 5
Solve for x
x = 1500 / 11 = 136 miles (rounded to the nearest mile)

At 11:00 a.m., John started driving along a highway at constant speed of 50 miles per hour. A quarter of an hour later, Jimmy started driving along the same highway in the same direction as John at the constant speed of 65 miles per hour. At what time will Jimmy catch up with John?
SolutionLet t be the number of hours, from 11:00 am, when Jimmy catches up with John. The time that Jimmy will have to drive to catch up with John is t - 1/4: he starts a quarter of an hour late. When Jimmy catches up with John, they would have traveled the same distance. Hence
50 t = 65 (t - 1/4)
Solve for t
50t = 65t - 65/4
t = 65/60 = 1.083 hours = 1hour and 5 minutes
Jim will catch up with John at
11:00 am + 1 hour , 5 minutes = 12:05 pm

A rectangle field has an area of 300 square meters and a perimeter of 80 meters. What are the length and width of the field?
SolutionLet L and W be the length and width of the rectangle. The area and the perimeter of the rectangle may be written as follows
L * W = 300 , area
2(L + W) = 80 or L + W = 40 , perimeter
Solve equation L + W = 40 for W
W = 40 - L
Substitute W by 40 - L in the equation L * W = 300
L(40 - L) = 300
Write the above equation in standard form
L2 - 40L + 300 = 0
(L - 30)(L - 10) = 0
Solve for L and calculate W
L = 30 and W = 40 - L = 10
L = 10 and W = 40 - 10
Assuming that the length is larger than the width, the rectangle has a length L = 30 meters and a width W = 10 meters

Find the area of a trapezoid whose parallel sides are 12 and 23 centimeters respectively.
SolutionThe area A of a trapezoid is given by
A = (1/2) height *(base1 + base2)
The two bases are given as 12 and 23 cm, however the height is not given and therefore there is not enough information to find the area of the trapezoid.

A rectangular garden in Mrs Dorothy's house has a length of 100 meters and a width of 50 meters. A square swimming pool is to be constructed inside the garden. Find the length of one side of the swimming pool if the remaining area (not occupied by the pool) is equal to one half the area of the rectangular garden.
Solutionlet x be the side of the swimming pool. If the area not covered by the swimming pool is half the rectangular garden, then the other half is covered by the swimming pool whose area is x2. Hence
x2 = (1/2) 100*50 = 2500
Solve for x
x = 50 meters , side of the swimming pool

ABC is an equilateral triangle with side length equal to 50 cm. BH is perpendicular to AC. MN is parallel to AC. Find the area of triangle BMN if the length of MN is equal to 12 cm.

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SolutionNote that since ABC is an equilateral triangle and BH is perpendicular to AC, HC = 50/2 = 25 cm. Also since MN and HC are parallel, the two triangles BMN and BHC are similar and the sizes of their sides are proportional. Hence
BN / BC = MN / HC = BM / BH = 12 / 25
The area of triangle BMN is given by
A = (1/2) BM * MN
The area of triangle BHC is given by
B = (1/2) BH * HC
Note the ratio of the two areas
A / B = [ (1/2) BM * MN ] / [ (1/2) BH * HC ] = (BM/BH) * (MN/HC) = (12/25)*(12/25) = (12/25)2let us now find BH using Pythagora's theorem.
BH = sqrt(502 - 252) = 25 sqrt(3)
The area B of triangle BHC is given by.
B = (1/2) BH * HC = (1/2) 25 sqrt(3) 25 = (1/2) 252 sqrt(3)
We now use the ratio A / B = (12/25)2 to find A the area of triangle BMN.
A = (12/25)2 * B = (12/25)2 * (1/2) 252 sqrt(3) = 72 sqrt(3) cm2

The height h of water in a cylindrical container with radius r = 5 cm is equal to 10 cm. Peter needs to measure the volume of a stone with a complicated shape and so he puts the stone inside the container with water. The height of the water inside the container rises to 13.2 cm. What is the volume of the stone in cubic cm?

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SolutionWe first find the volume of water without the stone.
V1 = 10 *(π * 52) = 250 π , where π = 3.14
The volume of water and stone is given by
V2 = 13.2 *(π * 52) = 330 π
The volume of the stone is given by
V2 - v1 = 330 π - 250 π = 80 π
= 251.1 cm3

In the figure below the square has all its vertices on the circle. The area of the square is equal to 400 square cm. What is the area of the circular shape?

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SolutionNote the size of the diagonal of the square is equal to the size of the diameter of the circle. The side x of the square is such that
x2 = 400 , hence x = 20 cm
The diagonal D of the square is found using Pythagora's theorem
D2 = x2 + x2 = 800
The area A of the circle is given by
A = π (D/2)2 = π D2 / 4 = 200 π cm3
= 628 cm3

The numbers 2 , 3 , 5 and x have an average equal to 4. What is x?
SolutionThe average of 2 , 3 , 5 and x is given by
(2 + 3 + 5 + x) / 4
and it is known and equal to 4. Hence
(2 + 3 + 5 + x) / 4 = 4
Solve the above equation for x to obtain
x = 6

The numbers x , y , z and w have an average equal to 25. The average of x , y and z is equal to 27. Find w.

1 liter is equal to 1 cubic decimeter and 1 liter of water weighs 1 kilogram. What is the weight of water contained in a cylindrical container with radius equal to 50 centimeters and height equal to 1 meter?

In the figure below triangle ABC is an isosceles right triangle. AM is perpendicular to BC and MN is perpendicular to AC. Find the ratio of the area of triangle MNC to the area of triangle ABC.

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SolutionABC is isosceles; hence
AB = AC
ABM and ACM have equal sides and one common side and equal angles are therefore congruent triangles. Hence the area of triangle AMC is half the area of ABC. In triangle MAC angles MAC and MCA are both equal to 45° hence triangle MAC is an isosceles right triangle and therefore trianles MAN and MCN are congruent (similar proof to above). Therefore area of triangle MNC is half the area of triangle AMC. Hence the ratio of the area of triangle MNC to the area of triangle ABC is equal to 1 / 4.

What are the dimensions of the square that has the perimeter and the area equal in value?
SolutionLet x be the side of the square. The perimeter 4x and the area x 2 are equal; hence
4 x = x 2Solve the above equation
4 x - x 2 = 0
x (4 - x) = 0
Two solution
x = 0 for which the square does not exist
x = 4 which is a square of side 4 units.

Find the dimensions of the rectangle that has a length 3 meters more that its width and a perimeter equal in value to its area?
SolutionLet L be the length and W be the width. "a length 3 meters more that its width" is written as
L = W + 3
The perimeter of the rectangle is given by
2 L + 2 W
The are of the rectangle is given by
L W
"perimeter equal in value to its area" is written as
2 L + 2 W = L W
Substitute L by 3 + W in the above equation to obtain
2 (W + 3)+ 2 W = (3+ W) W
Expand and write the equation in standard form
W 2 - W - 6 = 0
Solve the above quadratic equation to obtain
W = 3 and W = - 2
The width of a rectangle is positive hence W = 3 and
L = W + 3 = 6

The semicircle of area 50 π centimeters is inscribed inside a rectangle. The diameter of the semicircle coincides with the length of the rectangle. Find the area of the rectangle.
SolutionIf R is the radius of a semicircle; its area is known and is given by the formula
(1/2)πR 2 = 50 π
Solve the above equation for R
R 2 = 100 gives R = 10
The length L of the rectangle is equal to the diameter; hence
L = 2 R = 20 cm
The semicircle is inscribed inside the rectangle and therefore it radius is equal to width W of the rectangle; hence
W = R = 10 cm
The area of the rectangle is equal to
L × W = 20 × 10 = 200 cm 2

A triangle has an area of 200 cm2. Two sides of this triangle measure 26 and 40 cm respectively. Find the exact value of the third side.
SolutionThe area A of a triangle given its two sides a and b making an angle α is given by
A = (1/2) a b sin(α)
Use the cosine rule to find cos(α) as
cos (α) = (a 2 + b 2 - c 2) / 2 a b , c is the third side of the triangle.
Use the formula sin(α) = √(1- cos 2 (α)) to rewrite the formula for the area as
A = (1/2) a b sin(α) = (1/2) a b √(1- ((a 2 + b 2 - c 2) / 2 a b) 2)
Simplify to obtain
A = (1 / 4) √(4 a 2 b 2 - (a 2 + b 2 - c 2) 2)
Solve for c to obtain two solutions
c = √ [ a 2 + b 2 ~+mn~ √(4 a 2 b 2 - 16 A 2) ]
In this problem a = 26, b = 40 and A = 200. Substitute and solve to find two solutions
c1 = √ (356) = 2 √ (89) ; c2 = √ (4196) = 2 √ (1049)